Evaluate the following integrals: x^2 1-x^4 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{x}^2}{1-\text{x}^4}\ \text{dx}$

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Answer

Let $\text{I}=\int\frac{\text{x}^2}{1-\text{x}^4}\ \text{dx}$
we express
$\frac{\text{x}^2}{1-\text{x}^4}=\frac{\text{x}^2}{(1-\text{x})^2(1+\text{x})^2}$
$=\frac{\text{A}}{1-\text{x}^2}+\frac{\text{B}}{1+\text{x}^2}$
$\Rightarrow\text{x}^2=\text{A}(1+\text{x}^2)+\text{B}(1-\text{x})^2$
Equating the coefficient of x and constants, we get
1 = A - B and 0 = A + B or
$\text{A}=\frac{1}{2}$ and $\text{B}=-\frac{1}{2}$
$\therefore\text{I}=\int\Big(\frac{\frac{1}{2}}{1-\text{x}^2}+\frac{-\frac{1}{2}}{1+\text{x}^2}\Big)\text{dx}$
$=\frac{1}{2}\int\frac{1}{1-\text{x}^2}\ \text{dx}-\frac{1}{2}\int\frac{1}{1+\text{x}^2}\ \text{dx}$
$=\frac{1}{2}\times\frac{1}{2}\log\Big|\frac{1+\text{x}}{1-\text{x}}\Big|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
$=\frac{1}{4}\log\Big|\frac{1+\text{x}}{1-\text{x}}\Big|-\frac{1}{2}\tan^{-1}\text{t}+\text{C}$
Hence, $\int\frac{\text{x}^2}{1-\text{x}^4}\ \text{dx}=\frac{1}{4}\log\Big|\frac{1+\text{x}}{1-\text{x}}\Big|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$

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